Question

Write the equation of the parabola that passes through the points (2,0), (
–
1,0), and (3,8).

Answer 1

To find the equation of the parabola that passes through the points \((2,0)\), \((-1,0)\), and \((3,8)\), we can use the fact that the general form of a quadratic equation (parabola) is given by:

\[
y = ax^2 + bx + c
\]

Since \((2,0)\) and \((-1,0)\) are the x-intercepts of the parabola, we can express the quadratic in factored form:

\[
y = a(x - 2)(x + 1)
\]

We can expand this:

\[
y = a(x^2 - 2x + x - 2) = a(x^2 - x - 2)
\]

Thus, we have:

\[
y = ax^2 - ax - 2a
\]

Next, we need to use the third point \((3,8)\) to find the value of \(a\). Substitute \(x = 3\) and \(y = 8\):

\[
8 = a(3^2) - a(3) - 2a
\]

This simplifies to:

\[
8 = 9a - 3a - 2a
\]

Combining like terms, we get:

\[
8 = 4a
\]

Now, solve for \(a\):

\[
a = \frac{8}{4} = 2
\]

Now that we have \(a = 2\), we can substitute it back into the equation:

\[
y = 2(x^2 - x - 2)
\]

Expanding this gives:

\[
y = 2x^2 - 2x - 4
\]

Therefore, the equation of the parabola that passes through the points \((2,0)\), \((-1,0)\), and \((3,8)\) is:

\[
\boxed{y = 2x^2 - 2x - 4}
\]

Answer 2

Write your answer in the form y=a(x–p)(x–q), where a, p, and q are integers, decimals, or simplified fractions.